Let the archers be numbered $a_1, a_2, ... a_n$.
Let $p_i$ be the probability of archer $i$ making a shot.
Let $P_n(j, S)$ be the probability of archer $a_j$ winning out of the ordered set $S=\{i, ..., j, ... k\}$ with an initial size of $n$ where $a_j$ goes first.
General solution for $n=2$
For the following, we will compute a formula for 2 archers.
Lets say that we have two archers remaining, $a_i$ and $a_j$ and $a_i$ is to shoot first. To win, he can either kill $a_j$ on the first shot (happens with probability $p_i$), or if he misses (probability $1-p_i$) and survives the attempt by $a_j$ (probability $1-p_j$), then he can try again. This gives the following recursive formula:
$$P_n(i, \{i,j\}) = p_i + (1-p_i)(1-p_j)P(i,\{i,j\})$$
$$P_n(i, \{i,j\}) = \frac{p_i}{1-(1-p_i)(1-p_j)}$$
Since $p_k=\frac{k}{n}$, we can reduce this further.
$$P_n(i, \{i,j\}) =
\frac{\frac{i}{n}}{1-(1-\frac{i}{n})(1-\frac{j}{n})} =
\frac{i}{n \left( 1-\left( \frac{n-i}{n}\right) \left( \frac{n-j}{n} \right) \right)} =
\frac{i}{\frac{n^2-(n(i+j)+ij)}{n}} =
\frac{in}{ni+nj-ij}$$
Solution for $n=3$
As posted by @ghosts_in_the_code, there is a strategy for this scenario. When implemented, the following are the chances for survival.
Archer $a_3$
The first archer $a_1$ wastes his shot, and $a_2$ will aim at $a_3$. If he hits, then $a_3$ loses, but if he misses, then $a_3$ kills $a_2$ and must survive an attempt by $a_1$ to win. So, $a_3$ wins if both $a_2$ and $a_1$ miss.
$$P_3(3, \{1,2,3\})=(1-p_2)(1-p_1)=\frac{2}{9}$$
Archer $a_2$
The second archer will lose if he misses on his first shot at $a_3$ since $a_3$ will surely target him and kill him on his turn. Thus, $a_2$ must first kill $a_3$, and then survive $a_1$'s attempt. Then his chances are given in the formula above.
$$P_3(2, \{1,2,3\}) =
p_2 \times (1-p_1) \times P_3(2, \{1, 2\}) =
\frac{2}{3}\times \frac{2}{3}\times \frac{2 \times 3}{3 \times 2 + 3 \times 1 - 2 \times 1} =
\frac{8 \times 3}{3 \times 3 \times (6+3-2)} =
\frac{8}{21}$$
Archer $a_1$
The first archer can win in two ways:
- If $a_2$ kills $a_3$ ($p_2$) and then $a_1$ outlasts $a_2$ (formula above).
- If $a_3$ kills $a_2$ ($1- p_2$) and then $a_1$ hits $a_3$ on the first attempt ($p_1$). There will be only one attempt!
Using the formula we worked out before, we get the following:
$$P(1, \{1,2,3\}) =
p_2 \times P_3(1, \{1, 2\}) + (1-p_2) \times p_1 =
\frac{2}{3} \times \frac{3\times 1}{3 \times 1+ 3 \times 2 - 2 \times 1} + \left( 1-\frac{2}{3} \right) \times \frac{1}{3} =
\frac{2}{7} + \frac{1}{9} =
\frac{25}{63}$$
Summary
All these should add up to $1$, and they do.
$$\frac{25}{63} + \frac{8}{21} + \frac{2}{9} =
\frac{25 + (3 \times 8) + (7 \times 2)}{63} = \frac{25 + 24 + 14}{63} =
\frac{63}{63} = 1$$
Thus, somewhat counter-intuitively, the best archer has the worst chance of survival, and the worst archer has the best chance.
General Solution for $n=3$
We've done the specific solution for $a_1, a_2, a_3$, but what about for $a_i, a_j, a_k$? It is no longer obvious that the lowest archer shoot intentionally miss. For example, say $a_i$ and $a_j$ are equally dismal shots and $a_k$ is perfect. If $a_i$ wastes the shot and $a_j$ misses $a_k$, then $a_k$ will take out $a_j$. That leaves $a_i$ one shot to kill $a_k$. Thus, $a_i$'s chance of winning is $1-p_i$. If $a_j$ hits $a_k$ instead, then it is a duel between roughly equal players and $a_i$ has about a 50% chance.
Instead if $a_i$ aims for $a_k$ and hits, then although $a_j$ goes first in the duel, the chances are still pretty equal - much better than duelling $a_k$.
Thus, in this case, it is obvious that $a_i$ should shoot for $a_k$.
Thus we need to explore 2 strategies - one where the weakest wastes the shot, and the other where the weakest targets the strongest.
Weakest Targets Strongest
For the strongest to win, he needs to survive the first two shots. Then he will target $a_j$. If he is successful, then it is a two person game (and we have the general solution for that). If he fails, then it is the same probability as before.
$$P_n(k, \{i,j,k\}) = (1-p_i)(1-p_j)\Big( p_k \times P_n(k, \{i,k\}) + (1-p_k) \times P(k, \{i,j,k\}) \Big) =
\left(1-\frac{i}{n} \right) \left( 1-\frac{j}{n} \right) \left(
\frac{k}{n} \times \frac{nk}{nk+ni-ik} + \left( 1-\frac{k}{n} \right) \times P(k, \{i,j,k\})
\right)
$$
Solving this is rather tricky, and doesn't have a nice closed form that I can see.
So, for now, I am giving up on this. Perhaps we can come back to it later, but I think I am taking the wrong approach.
